Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Fermat’s Last Theorem over Quadratic ImaginaryFields of Class Number one. An overview of the

difficulties

George Turcas

University of Warwick

7th of June, 2017Journees Algophantiennes Bordelaises

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Introduction

ap + bp + cp = 0, abc 6= 0 Ea,b,c : Y 2 = X (X − ap)(X + bp)

Jarvis and Meekin (2004) FLT over Q(√

2)

Freitas and Siksek (2014) FLT over some real quadratic #fields

Key to this successes is the extraordinary progress inmodularity over totally real fields due to the efforts ofBarnett-Lamb, Breuil, Diamond, Gee, Geraghty, Kisin,Skinner, Taylor, Wiles and others.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Our understanding of modularity (or automorphy) in the setting ofgeneral number fields is highly conjectural. In recent work,assuming two conjectures, Sengun and Siksek manage to replicateto some extent the aforementioned successes for Fermat’s equationover general number fields.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Conjecture ((1) Serre’s modularity conjecture)

This conjecture predicts the existence of a weight 2 mod peigenform over K which is associated to every odd, absolutelyirreducible continous representation ρ : GK → GL2(Fp) that has avery special property at every prime p|p of K .

Conjecture (2)

This is a conjecture in the Langlands Programme which says thatevery weight 2 newform (for GL2) over K with integer Heckeeigenvalues has an associated elliptic curve over K or a fake ellipticcurve over K .

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Notation

K - a number field

ZK - the ring of integers of K

We consider the Fermat equation with prime exponent p ∈ Z

xp + yp + zp = 0, (1)

with x , y , z ∈ K .

A solution (a, b, c) ∈ K 3 to (1) is called non-trivial if abc 6= 0.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Theorem (Sengun, Siksek -2016)

Let K = Q(√−d) be an imaginary quadratic field where d is a

squarefree positive integer satisfying −d = 2 or 3 (mod 4).Assume that conjectures 1 and 2 hold. Then, there exists aconstant BK such that for p > BK , Fermat’s equation

xp + yp + zp = 0

does not have non-trivial solutions in K .

Remark

The constant BK is ineffective.

From now on, let K be an imaginary quadratic field of classnumber 1, i.e. K = Q(

√−d), where d is one of 1, 2, 3, 7, 11, 19,

43, 67 or 163.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Theorem (Sengun, Siksek -2016)

Let K = Q(√−d) be an imaginary quadratic field where d is a

squarefree positive integer satisfying −d = 2 or 3 (mod 4).Assume that conjectures 1 and 2 hold. Then, there exists aconstant BK such that for p > BK , Fermat’s equation

xp + yp + zp = 0

does not have non-trivial solutions in K .

Remark

The constant BK is ineffective.

From now on, let K be an imaginary quadratic field of classnumber 1, i.e. K = Q(

√−d), where d is one of 1, 2, 3, 7, 11, 19,

43, 67 or 163.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Theorem (Sengun, Siksek -2016)

Let K = Q(√−d) be an imaginary quadratic field where d is a

squarefree positive integer satisfying −d = 2 or 3 (mod 4).Assume that conjectures 1 and 2 hold. Then, there exists aconstant BK such that for p > BK , Fermat’s equation

xp + yp + zp = 0

does not have non-trivial solutions in K .

Remark

The constant BK is ineffective.

From now on, let K be an imaginary quadratic field of classnumber 1, i.e. K = Q(

√−d), where d is one of 1, 2, 3, 7, 11, 19,

43, 67 or 163.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Let (a, b, c) ∈ K 3 be a non-trivial solution to the Fermat equation.We can scale (a, b, c) such that the triple is integral and asCl(K ) = 1, we can suppose that a, b, c are coprime. We associateto (a, b, c) the following Frey curve

E = Ea,b,c : Y 2 = X (X − ap)(X + bp).

Let ρ = ρE ,p : GK → Aut(E [p]) ∼= GL2(Fp) be the representationinduced by the action of GK on the p-torsion E [p].

Plan

Prove that ρ satisfies the hypothesis required for Serre’s modularityconjecture. Get an weight 2 eigenform for K at a level that doesnot depend on p or a, b, c . Compute the space of those eigenformsand hope for a contradiction. This will prove that (a, b, c) ∈ K 3

does not exist.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Let (a, b, c) ∈ K 3 be a non-trivial solution to the Fermat equation.We can scale (a, b, c) such that the triple is integral and asCl(K ) = 1, we can suppose that a, b, c are coprime. We associateto (a, b, c) the following Frey curve

E = Ea,b,c : Y 2 = X (X − ap)(X + bp).

Let ρ = ρE ,p : GK → Aut(E [p]) ∼= GL2(Fp) be the representationinduced by the action of GK on the p-torsion E [p].

Plan

Prove that ρ satisfies the hypothesis required for Serre’s modularityconjecture. Get an weight 2 eigenform for K at a level that doesnot depend on p or a, b, c . Compute the space of those eigenformsand hope for a contradiction. This will prove that (a, b, c) ∈ K 3

does not exist.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

E : Y 2 = X (X − ap)(X + bp)

The determinant of ρ is the mod p cyclotomic character;

c4 = 24(b2p − apcp) and ∆ = 24(abc)2p

for any prime q of ZK that does not lie above 2, the givenmodel for E is minimal at q and E is semistable at q.

for such q, we have p|vq(∆)

ρ is unramified at any prime that does not lie above 2 and p.

ρ is finite flat at q if q|pBy the above, the Serre conductor N of ρ belongs to a finiteset that depends only on the field K .

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

E : Y 2 = X (X − ap)(X + bp)

The determinant of ρ is the mod p cyclotomic character;

c4 = 24(b2p − apcp) and ∆ = 24(abc)2p

for any prime q of ZK that does not lie above 2, the givenmodel for E is minimal at q and E is semistable at q.

for such q, we have p|vq(∆)

ρ is unramified at any prime that does not lie above 2 and p.

ρ is finite flat at q if q|pBy the above, the Serre conductor N of ρ belongs to a finiteset that depends only on the field K .

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

E : Y 2 = X (X − ap)(X + bp)

The determinant of ρ is the mod p cyclotomic character;

c4 = 24(b2p − apcp) and ∆ = 24(abc)2p

for any prime q of ZK that does not lie above 2, the givenmodel for E is minimal at q and E is semistable at q.

for such q, we have p|vq(∆)

ρ is unramified at any prime that does not lie above 2 and p.

ρ is finite flat at q if q|pBy the above, the Serre conductor N of ρ belongs to a finiteset that depends only on the field K .

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

E : Y 2 = X (X − ap)(X + bp)

The determinant of ρ is the mod p cyclotomic character;

c4 = 24(b2p − apcp) and ∆ = 24(abc)2p

for any prime q of ZK that does not lie above 2, the givenmodel for E is minimal at q and E is semistable at q.

for such q, we have p|vq(∆)

ρ is unramified at any prime that does not lie above 2 and p.

ρ is finite flat at q if q|pBy the above, the Serre conductor N of ρ belongs to a finiteset that depends only on the field K .

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

E : Y 2 = X (X − ap)(X + bp)

The determinant of ρ is the mod p cyclotomic character;

c4 = 24(b2p − apcp) and ∆ = 24(abc)2p

for any prime q of ZK that does not lie above 2, the givenmodel for E is minimal at q and E is semistable at q.

for such q, we have p|vq(∆)

ρ is unramified at any prime that does not lie above 2 and p.

ρ is finite flat at q if q|pBy the above, the Serre conductor N of ρ belongs to a finiteset that depends only on the field K .

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

E : Y 2 = X (X − ap)(X + bp)

The determinant of ρ is the mod p cyclotomic character;

c4 = 24(b2p − apcp) and ∆ = 24(abc)2p

for any prime q of ZK that does not lie above 2, the givenmodel for E is minimal at q and E is semistable at q.

for such q, we have p|vq(∆)

ρ is unramified at any prime that does not lie above 2 and p.

ρ is finite flat at q if q|p

By the above, the Serre conductor N of ρ belongs to a finiteset that depends only on the field K .

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

E : Y 2 = X (X − ap)(X + bp)

The determinant of ρ is the mod p cyclotomic character;

c4 = 24(b2p − apcp) and ∆ = 24(abc)2p

for any prime q of ZK that does not lie above 2, the givenmodel for E is minimal at q and E is semistable at q.

for such q, we have p|vq(∆)

ρ is unramified at any prime that does not lie above 2 and p.

ρ is finite flat at q if q|pBy the above, the Serre conductor N of ρ belongs to a finiteset that depends only on the field K .

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

To apply Conjecture (1) to the mod p representation ρ of the Freycurve E , we need to prove that ρ is absolutely irreducible.For the exposition, we specialise further to the case K = Q(i),where 2ZK = a2 ramifies. Here a = (1 + i)ZK .

Theorem

For p > 17, ρ is irreducible.

Sketch of proof. Suppose

ρ ∼(θ ∗0 θ′

),

where θ, θ′ : GK → F∗p are characters, such that θθ′ = χp.It is easy to show that if q - p is a prime of additive reduction, then

vq(Nθ) = vq(Nθ′) =1

2vq(N ).

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

i) Suppose that p is coprime to Nθ or Nθ′ . By replacing E with ap isogenous curve if necessary we can assume that p is coprime toNθ.

finitely many choices for Nθ;

θ is the character of a ray class group for which we havefinitely many candidates;

in this very fortunate situation θ has order 1 or 2;

this gives a p torsion point for E over K , or a degree 4extension of Q;

possible prime torsion over a quartic extension of Q weredetermined by Derickx, Kamienny, Stein and Stoll and it isknow that p ≤ 17.

ii) p is not coprime to Nθ nor Nθ′ . Using work of David and ideasof Freitas and Siksek, we can get a contradiction.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Lemma (Freitas, Siksek)

Let a be a prime of ZK that lies above 2 and has inertial degree 1.If p > 4va(2), then(i) E has potentially multiplicative reduction at a;(ii) p|#ρ(Ia), where Ia denotes the inertia subgroup of GK at a.

E has potentially multiplicative reduction at a and the imageof ρ contains an element of order p;

any irreducible subgroup of GL2(Fp) that has elements oforder p contains SL2(Fp);

K ∩Q(ζp) = Q, therefore det(ρ) = χp is surjective, whichimplies that ρ is surjective.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

In the proof of Fermat’s Last Theorem over Q, Serre’smodularity conjecture predicts that the representation ρcomes from an eigenform of weight 2 and level 2.

Over a general number field, Serre’s modularity conjecturepredicts that the representation comes from a mod peigenform of a certain weight and level.

For K = Q(i), since ρ satisfies all the hypothesis of theSerre’s conjecture, the latter predicts that there exists anon-trivial (cuspidal) weight 2 mod p eigenform over K oflevel N (= the Serre conductor of ρ) associated to ourresidual Galois representation.

here N is supported only on S = {a} and it is at most a8.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

In the proof of Fermat’s Last Theorem over Q, Serre’smodularity conjecture predicts that the representation ρcomes from an eigenform of weight 2 and level 2.

Over a general number field, Serre’s modularity conjecturepredicts that the representation comes from a mod peigenform of a certain weight and level.

For K = Q(i), since ρ satisfies all the hypothesis of theSerre’s conjecture, the latter predicts that there exists anon-trivial (cuspidal) weight 2 mod p eigenform over K oflevel N (= the Serre conductor of ρ) associated to ourresidual Galois representation.

here N is supported only on S = {a} and it is at most a8.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Over Q (and totally real number fields), mod p eigenforms aresimply mod p reductions of complex eigenforms and we don’tsee the phrase ”mod p eigenform” in the statement of Serre’sconjecture.

PGL2(K ) acts on H3 via the embedding GL2(K ) ↪→ GL2(C).

Y0(N ) := Γ0(N )\H3,

where Γ0(N ) is the usual congruence subgroup of PGL2(ZK ).H i (Y0(N ),C) and H i (Y0(N ),Fp) have Hecke operators onthem.

The obstruction in lifting mod p eigenforms to complexeigenforms is given by the torsion in H2(Y0(N ),Z).

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

I used an algorithm of Sengun for computing the torsion and Ichecked that for K and N as presented, H2(Y0(N ),Z) doesnot have p torsion, therefore ρ comes from a complexeigenform of level N .

The predicted complex eigenform is cuspidal and using theMAGMA (Bosma at all) implementation of an algorithm ofGunnels we can compute the cuspidal part of this space.

We see that the aforementioned cuspidal spaces are emptyand therefore, we obtain the following theorem.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Theorem

Assume conjecture 1 holds for K = Q(i). Then, for p > 17,Fermat’s equation with prime exponent

xp + yp + zp = 0

does not have any solution (a, b, c) ∈ Q(i)3 such that abc 6= 0.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Remark

We were very lucky in this case!

For some of the number fields that we considered, we findweight 2 complex eigenforms over K . When K has a realembedding, such an eigenform over K (with rational Heckeeigenvalues) corresponds conjecturally to an elliptic curve overK .

If K is totally complex, the eigenform does sometimescorrespond to a fake elliptic curve. Sengun and Siksekobserved that images ρ(Ia) of inertia at primes a above 2 thathave residual degree 1 are incompatible with images of inertiafor fake elliptic curves.

2 ramifies in Q(i),Q(√−2), splits in Q(

√−7) and stays inert

in the other 6 fields of interest for us.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Thank you very much for listening!

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

It is time to give a more precise statement of Conjecture (1)

Conjecture (1)

Let ρ : GK → GL2(Fp) be an odd, irreducible, continuousrepresentation with Serre conductor N (prime-to-p part of its Artinconductor) and trivial character (prime-to-p part of det(ρ)).Assume that p is unramified in K and that ρ |GKp

arises from a

finite-flat group scheme over ZKp for every prime p|p. Then thereis a (weight 2) mod p eigenform θ over K of level N such that forall primes q coprime to pN, we have

Tr(ρ(Frobq)) = θ(Tq).

Remark

We say that ρ is odd if the determinant of every complexconjugation is −1. In our case, K is totally complex and we regardρ automatically as odd.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

It is time to give a more precise statement of Conjecture (1)

Conjecture (1)

Let ρ : GK → GL2(Fp) be an odd, irreducible, continuousrepresentation with Serre conductor N (prime-to-p part of its Artinconductor) and trivial character (prime-to-p part of det(ρ)).Assume that p is unramified in K and that ρ |GKp

arises from a

finite-flat group scheme over ZKp for every prime p|p. Then thereis a (weight 2) mod p eigenform θ over K of level N such that forall primes q coprime to pN, we have

Tr(ρ(Frobq)) = θ(Tq).

Remark

We say that ρ is odd if the determinant of every complexconjugation is −1. In our case, K is totally complex and we regardρ automatically as odd.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

It is time to give a more precise statement of Conjecture (1)

Conjecture (1)

Let ρ : GK → GL2(Fp) be an odd, irreducible, continuousrepresentation with Serre conductor N (prime-to-p part of its Artinconductor) and trivial character (prime-to-p part of det(ρ)).Assume that p is unramified in K and that ρ |GKp

arises from a

finite-flat group scheme over ZKp for every prime p|p. Then thereis a (weight 2) mod p eigenform θ over K of level N such that forall primes q coprime to pN, we have

Tr(ρ(Frobq)) = θ(Tq).

Remark

We say that ρ is odd if the determinant of every complexconjugation is −1. In our case, K is totally complex and we regardρ automatically as odd.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

GL2(K ) acts on the hyperbolic 3-space H3 via the embeddingGL2(K ) ↪→ GL2(K ⊗ R) ' GL2(C). Fix an ideal N ⊆ ZK anddefine the compact open subgroup

U0(N) :=

{γ ∈ GL2(ZK ) : γ ≡

(∗ ∗0 ∗

)mod N

}.

The locally symmetric space

Y0(N) = GL2(K )\((

GL2(AfK )/U0(N)

)×H3

)is in this particular case just a Riemannian 3-fold

Y0(N) = Γ0(N)\H3,

where Γ0(N) is the usual congruence subgroup Γ0(N) of themodular group GL2(ZK ).

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

For i ∈ {1, 2} consider the i-th cohomology group H i (Y0(N),C).For any prime q coprime to the level N, we can construct a linearendomorphism Tq of H i (Y0(N),C), called a Hecke operator. Let

T(i)C (N) be the commutative Z-algebra generated by these Hecke

operators inside the endomorphism algebra of H i (Y0(N),C).A complex eigenform f over K of degree i and level N is a ring

homomorphism f : T(i)C (N)→ C.

values of f generate a # field Qf .

f is trivial if f (Tq) = Norm(q) + 1, ∀q - Nf , g are equivalent if f (Tq) = g(Tq), for almost all q. f , gare allowed to have different degrees and levels.

f is called new if it is not equivalent to one whose level is aproper divisor of N.

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

If p is a rational prime unramified in K and coprime to the level,H i(Y0(N),Fp

)also comes equipped ith Hecke operators Tq where

q - pN. They form an algebra T(i)

Fp. A (weight 2) mod p

eigenform f over K of degree i and level N is a ring

homomorphism f : T(i)

Fp(N)→ Fp.

Definition

We say that a mod p eigenform θ, of level N, lifts to a complexeigenform if there exists a complex eigenform f , of the same degreeand level and a prime ideal p of Qf over p such that for everyprime q of K coprime to pN we have θ(Tq) = f (Tq) (mod p).

Introduction The Frey Curve and its mod p Galois rep. Surjectivity of ρ Conclusions Further difficulties

Remark

Not all eigenforms lift to complex ones. The obstruction to liftingis given by p-torsion in the integral cohomology.

The long exact sequence associated to te multiplication by p shortexact sequence 0→ Z p−→ Z→ Fp → 0 gives rise to the followingshort exact sequence

0→ H i (Y0(N),Z)⊗Fp → H i (Y0(N),Fp)→ H i+1 (Y0(N),Z) [p]→ 0.

The existence of an eigenform (complex or mod p) is equivalent tothe existence of a class in the corresponding cohomology groupthat is a simultaneous eigenvector for the Hecke operators suchthat its eigenvalues match the values of the form.With this interpretation, one can use lifting lemmas of Ash andStevens to deduce that every mod p eigenform of degree i lifts to acomplex one when H i+1(Y0(N),Z) has no p-torsion.